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Non Lebesgue integrability

  1. Sep 1, 2011 #1
    Hello all, can someone please direct me towards an argument proving the Lebesgue integral from 0 to infinity of sin x / x does not exist?
    Many thanks
     
  2. jcsd
  3. Sep 1, 2011 #2

    micromass

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    We need to show that

    [tex]\int_0^\infty{\left|\frac{\sin(x)}{x}\right|dx}=+\infty[/tex]

    For this, we set

    [tex]J_k=\int_{k\pi}^{(k+1)\pi}{\left|\frac{\sin(x)}{x}\right|dx}[/tex]

    Change the variables: [itex]y=x-k\pi[/itex] to obtain

    [tex]J_k=\int_0^\pi{\frac{\sin(y)}{y+k\pi}dy}[/tex]

    From [itex]y+k\pi\leq (k+1)\pi[/itex] follows

    [tex]J_k\geq \frac{1}{(k+1)\pi}\int_0^\pi{\sin(x)dx}=\frac{2}{(k+1)\pi}[/tex]

    Thus

    [tex]\int_0^{+\infty}{\left|\frac{\sin(x)}{x}\right|dx}\geq \sum_{k=0}^{+\infty}{J_k}=+\infty[/tex]
     
  4. Sep 2, 2011 #3
    Micromass,

    many thanks for thr neat proof.
    The fact that this function is no Lebesgue integrable but is it Riemann integrable in the improper sense is the most puzzling to me.
    Can you please maybe attempt an heuristic explanation too?

    Wikipedia says that "from the point of view of meausre theory the integral is like infty - infty", which I do not understand at all.

    Also, on the integrabiliyt of thre function 1/x. In this thread https://www.physicsforums.com/archive/index.php/t-276377.htmlvigVig offers a proof, but i really do not get why would phi(x) converge to infty, and i did not manage to clarify.
    Many thanks for your help

    Muzialis
     
  5. Sep 2, 2011 #4

    mathman

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    sinx/x is integrable only in the sense that you can integrate it from 0 to X and then let X become infinite. The integral from 0 to X exists in both Riemann and Lebesgue definitions.
     
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